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Chapter 20: Problem 2

The specific heat of water is \(4184 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\). By how many joules does the internal energy of \(50 \mathrm{~g}\) of water change as it is heated from \(21{ }^{\circ} \mathrm{C}\) to \(37{ }^{\circ} \mathrm{C}\) ? Assume that the expansion of the water is negligible. The heat added to raise the temperature of the water is $$ \Delta Q=c m \Delta T=(4184 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K})(0.050 \mathrm{~kg})\left(16^{\circ} \mathrm{C}\right)=3.4 \times 10^{3} \mathrm{~J} $$ Notice that \(\Delta T\) in Celsius is equal to \(\Delta T\) in kelvin. If we ignore the slight expansion of the water, no work was done on the surroundings and so \(\Delta W=0 .\) Then, the first law, \(\Delta Q=\Delta U+\Delta W\), tells us that $$ \Delta U=\Delta Q=3.4 \mathrm{~kJ} $$

Short Answer

Expert verified
The internal energy change is 3347.2 J.

Step by step solution

01

Evaluate the Given Values

Identify the given values in the problem: the specific heat capacity of water is \(c = 4184 \text{ J/kg} \cdot \text{K}\), the mass of water \(m = 50 \text{ g} = 0.050 \text{ kg}\), the initial temperature \(T_i = 21 ^{\circ} \text{C}\), and the final temperature \(T_f = 37 ^{\circ} \text{C}\).
02

Determine the Change in Temperature

Calculate the change in temperature \(\Delta T\) by subtracting the initial temperature from the final temperature: \(\Delta T = T_f - T_i = 37 ^{\circ} \text{C} - 21 ^{\circ} \text{C} = 16 ^{\circ} \text{C}\). Note that \(\Delta T\) in Celsius is the same as in Kelvin.
03

Apply the Formula for Heat Change

Use the formula \(\Delta Q = c \cdot m \cdot \Delta T\) to calculate the heat added. Substitute the known values: \(\Delta Q = 4184 \text{ J/kg} \cdot \text{K} \times 0.050 \text{ kg} \times 16 \text{ K} = 3347.2 \text{ J}\).
04

Relate Heat Added to Internal Energy Change

According to the first law of thermodynamics, \(\Delta Q = \Delta U + \Delta W\). Since \(\Delta W = 0\) (no work is done due to negligible expansion), we have \(\Delta U = \Delta Q\).
05

Express Final Result for Internal Energy Change

Therefore, the change in internal energy \(\Delta U = 3347.2 \text{ J}\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

First Law of Thermodynamics
The First Law of Thermodynamics is a fundamental principle in physics that helps us understand how energy moves and changes form. Essentially, it states that energy cannot be created or destroyed; it can only change forms. This law is often connected to a simple but powerful equation:
  • \( \Delta Q = \Delta U + \Delta W \)
This equation tells us that the change in energy added to a system by heating (\( \Delta Q \)) is equal to the change in internal energy (\( \Delta U \)) plus the work done by the system (\( \Delta W \)). The equation is like a balance sheet for energy. It helps us track where energy goes within a system, whether it stays inside as internal energy or escapes as work.
For this exercise, let's assume a system where water heats up. The work done (\( \Delta W \)) is zero because we are ignoring expansion. Thus, any energy we add as heat directly increases the water's internal energy. This helps us solve the problem more straightforwardly, focusing purely on the heat energy that changes the water's temperature.
Internal Energy Change
Internal energy is a fascinating concept. It refers to the total energy contained within a system. This energy arises from all the particles that make up the system, depending on their speed, type, and arrangement.
When you add heat to water, its internal energy changes, as the heat energizes the water molecules, making them move faster. In this exercise, we determined that the water's internal energy changed by about 3347.2 joules once it was heated.
  • Internal energy change formula: \( \Delta U = \Delta Q - \Delta W \)
  • Since \( \Delta W = 0 \), we simplify it: \( \Delta U = \Delta Q \)
By knowing the specific heat capacity of water, the quantity of water you're working with, and the temperature change, you can accurately calculate this change in internal energy. This is crucial information, especially in fields involving thermodynamics or heat exchange systems.
Temperature Change
Temperature change is another key topic in thermodynamics. It measures how much hotter or colder a substance becomes and connects directly to the energy exchange within the system. In practical terms, the temperature change is what you notice when you heat your tea or when a metal rod heats up on the stove.
In this problem, we calculated the water's temperature change from \( 21^{\circ} \text{C} \) to \( 37^{\circ} \text{C} \). That's a change of \( 16^{\circ} \text{C} \). Because Celsius and Kelvin have the same increment size, this change is equivalent to \( 16 \text{ K} \).
  • Temperature change formula: \( \Delta T = T_f - T_i \)
Understanding the temperature change is crucial because it directly affects how much heat you'll need to adjust that temperature, particularly when you employ the specific heat capacity. This value defines how much energy you need to change the temperature of a unit mass by one degree. Specific heat is what makes water such an effective coolant, as it takes in a lot of energy for only a small change in temperature.

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Most popular questions from this chapter

As \(3.0\) liters of ideal gas at \(27^{\circ} \mathrm{C}\) is heated, it expands at a constant pressure of \(2.0 \mathrm{~atm}\). How much work is done by the gas as its temperature is changed from \(27^{\circ} \mathrm{C}\) to \(227^{\circ} \mathrm{C}\) ?

The temperature of \(5.00 \mathrm{~kg}\) of \(\mathrm{N}_{2}\) gas is raised from \(10.0{ }^{\circ} \mathrm{C}\) to \(130.0{ }^{\circ} \mathrm{C}\). If this is done at constant volume, find the increase in internal energy \(\Delta U .\) Alternatively, if the same temperature change now occurs at constant pressure determine both \(\Delta V\) and the external work \(\Delta W\) done by the gas. For \(\mathrm{N}_{2}\) gas, \(c_{v}=0.177 \mathrm{cal} / \mathrm{g} \cdot{ }^{\circ} \mathrm{C}\) and \(c_{p}=0.248 \mathrm{cal} / \mathrm{g} \cdot{ }^{\circ} \mathrm{C}\). If the gas is heated at constant volume, then no work is done during the process. In that case \(\Delta W=0\), and the first law tells us that \((\Delta Q)_{v}=\Delta U .\) Because \((\Delta Q)_{v}=c_{v} m \Delta T\), $$\Delta U=(\Delta Q)_{v}=\left(0.177 \mathrm{cal} / \mathrm{g} \cdot{ }^{\circ} \mathrm{C}\right)(5000 \mathrm{~g})\left(120^{\circ} \mathrm{C}\right)=106 \mathrm{kcal}=443 \mathrm{~kJ}$$ The temperature change is a manifestation of the internal energy change. When the gas is heated \(120^{\circ} \mathrm{C}\) at constant pressure, the same change in internal energy occurs. In addition, however, work is done. The first law then becomes $$(\Delta Q)_{p}=\Delta U+\Delta W=443 \mathrm{~kJ}+\Delta W$$ But $$\begin{aligned} (\Delta Q)_{p} &=c_{p} m \Delta T=\left(0.248 \mathrm{cal} / \mathrm{g} \cdot{ }^{\circ} \mathrm{C}\right)(5000 \mathrm{~g})\left(120{ }^{\circ} \mathrm{C}\right) \\ &=149 \mathrm{kcal}=623 \mathrm{~kJ} \end{aligned}$$ $$ \text { Hence } \quad \Delta W=(\Delta Q)_{p}-\Delta U=623 \mathrm{~kJ}-443 \mathrm{~kJ}=180 \mathrm{~kJ} $$

Find \(\Delta W\) and \(\Delta U\) for a \(6.0-\mathrm{cm}\) cube of iron as it is heated from \(20^{\circ} \mathrm{C}\) to \(300{ }^{\circ} \mathrm{C}\) at atmospheric pressure. For iron, \(c=0.11 \mathrm{cal} / \mathrm{g} \cdot{ }^{\circ} \mathrm{C}\) and the volume coefficient of thermal expansion is \(3.6 \times 10^{-5}{ }^{\circ} \mathrm{C}^{-1}\). The mass of the cube is \(1700 \mathrm{~g}\). Given that \(\Delta T=300^{\circ} \mathrm{C}-20^{\circ} \mathrm{C}=280^{\circ} \mathrm{C}\) $$ \Delta Q=c m \Delta T=\left(0.11 \mathrm{cal} / \mathrm{g} \cdot{ }^{\circ} \mathrm{C}\right)(1700 \mathrm{~g})\left(280{ }^{\circ} \mathrm{C}\right)=52 \mathrm{kcal} $$ To find that the work done by the expansion of the cube, we need to determine \(\Delta V\). The volume of the cube is \(V=(6.0 \mathrm{~cm})^{3}=216 \mathrm{~cm}^{3} .\) Using \((\Delta V) / V=\beta \Delta T\), $$ \Delta V=V \beta \Delta T=\left(216 \times 10^{-6} \mathrm{~m}^{3}\right)\left(3.6 \times 10^{-5}{ }^{\circ} \mathrm{C}^{-1}\right)\left(280{ }^{\circ} \mathrm{C}\right)=2.18 \times 10^{-6} \mathrm{~m}^{3} $$ Then, assuming atmospheric pressure to be \(1.0 \times 10^{5} \mathrm{~Pa}\), $$ \Delta W=P \Delta V=\left(1.0 \times 10^{5} \mathrm{~N} / \mathrm{m}^{2}\right)\left(2.18 \times 10^{-6} \mathrm{~m}^{3}\right)=0.22 \mathrm{~J} $$ But the First Law tells us that $$ \begin{aligned} \Delta U &=\Delta Q-\Delta W=(52000 \mathrm{cal})(4.184 \mathrm{~J} / \mathrm{cal})-0.22 \mathrm{~J} \\ &=218000 \mathrm{~J}-0.22 \mathrm{~J} \approx 2.2 \times 10^{5} \mathrm{~J} \end{aligned} $$ Notice how very small the work of expansion against the atmosphere is in comparison to \(\Delta U\) and \(\Delta Q .\) Often \(\Delta W\) can be neglected when dealing with liquids and solids.

Twenty cubic centimeters of monatomic gas at \(12^{\circ} \mathrm{C}\) and \(100 \mathrm{kPa}\) is suddenly (and adiabatically) compressed to \(0.50 \mathrm{~cm}^{3}\). Assume that we are dealing with an ideal gas. What are its new pressure and temperature? For an adiabatic change involving an ideal gas, \(P_{1} V_{1}^{\gamma}=P_{2} V_{2}^{\gamma}\) where \(\gamma=1.67\) for a monatomic gas. Hence, $$P_{2}=P_{1}\left(\frac{V_{1}}{V_{2}}\right)^{\gamma}=\left(1.00 \times 10^{5} \mathrm{~N} / \mathrm{m}^{2}\right)\left(\frac{20}{0.50}\right)^{1.67}=4.74 \times 10^{7} \mathrm{~N} / \mathrm{m}^{2}=47 \mathrm{MPa}$$ To find the final temperature, we could use \(P_{1} V_{1} / T_{1}=P_{2} V_{2} / T_{2}\). Instead, let us use $$T_{1} V_{1}^{\gamma-1}=T_{2} V_{2}^{\gamma-1}$$ or $$T_{2}=T_{1}\left(\frac{V_{1}}{V_{2}}\right)^{\gamma-1}=(285 \mathrm{~K})\left(\frac{20}{0.50}\right)^{0.67}=(285 \mathrm{~K})(11.8)=3.4 \times 10^{3} \mathrm{~K}$$ As a check, $$ \begin{array}{c} \frac{P_{1} V_{1}}{T_{1}}=\frac{P_{2} V_{2}}{T_{2}} \\ \frac{\left(1 \times 10^{5} \mathrm{~N} / \mathrm{m}^{2}\right)\left(20 \mathrm{~cm}^{3}\right)}{285 \mathrm{~K}}=\frac{\left(4.74 \times 10^{7} \mathrm{~N} / \mathrm{m}^{2}\right)\left(0.50 \mathrm{~cm}^{3}\right)}{3370 \mathrm{~K}} \\ 7000=7000 \swarrow \end{array} $$

(a) Compute \(c_{v}\) for the monatomic gas argon, given \(c_{p}=0.125 \mathrm{cal} / \mathrm{g} \cdot{ }^{\circ} \mathrm{C}\) and \(\gamma=1.67 .(b)\) Compute \(c_{p}\) for the diatomic gas nitric oxide (NO), given \(c_{v}=0.166 \mathrm{cal} / \mathrm{g} \cdot{ }^{\circ} \mathrm{C}\) and \(\gamma=1.40\).
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